What Is the Minimum Stopping Distance for a Car Traveling at 50 m/s?

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The discussion centers on calculating the minimum stopping distance for a car traveling at 50 m/s, building on a previous example where a car at 30 m/s had a stopping distance of 60 m, including a 0.5-second reaction time. Participants emphasize the importance of incorporating the reaction time into the calculations, which affects the total stopping distance. The initial deceleration must be determined from the first scenario, and then applied to the new speed while adding the distance covered during the reaction time. Confusion arises regarding how to integrate these elements into the stopping distance formula. Ultimately, the conversation highlights the need for clarity in applying physics principles to real-world driving scenarios.
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The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver's reaction time of 0.50 s.

(a) What is the minimum stopping distance for the same car traveling at a speed of 50 m/s?


Vf^2 = Vi^2 +2ax


So i would be solving for X here right? so the car is not accelerating do i plug in 0 for a?

This is what i have so far

0=50^2+2(a)(x) where does the .5 reaction time come into play?
 
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jo3jo3520 said:
The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver's reaction time of 0.50 s.

(a) What is the minimum stopping distance for the same car traveling at a speed of 50 m/s?


Vf^2 = Vi^2 +2ax


So i would be solving for X here right? so the car is not accelerating do i plug in 0 for a?

This is what i have so far

0=50^2+2(a)(x) where does the .5 reaction time come into play?

You're right that the .5sec reaction time needs to be accounted for.

But you need to first account for it from the first measurements.

The Total distance traveled to stopping included .5*(30m/s). Then the total change in velocity occurred over that smaller distance - not including reaction time.

Now for the 50m/s case how long to stop if you use the deceleration from the first measured stop? From that distance you then need to add the (50m/s)*.5sec to get your total distance.
 
ok so .5*30 = 15 but I am confused on what to do with that? you said "Then the total change in velocity occurred over that smaller distance - not including reaction time." ?
 
am i supposed to do 30/15? and where in the equation do i plug this number in?
 
help anyone?
 
jo3jo3520 said:
ok so .5*30 = 15 but I am confused on what to do with that? you said "Then the total change in velocity occurred over that smaller distance - not including reaction time." ?

So the car deceleration alone was only over 45m and not 60m then wasn't it? Now figure the deceleration performance of just the car.

You have Velocity and distance, so find acceleration "a" that gives you that deceleration over 45m.
 
Either I am just really dumb or not understanding but i give up. thank you for your help
 

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